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VashaNatasha [74]
3 years ago
8

A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L

of a 33.0 mM solution?
Chemistry
1 answer:
slega [8]3 years ago
5 0

Answer:

5.41 g

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

Moles=0.033 \times {0.45}\ moles

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.01485\ g= \frac{Mass}{342.39\ g/mol}

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

<u>Answer -  5.41 g</u>

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